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Let $\Psi_n$ denote the $n$-th division polynomial associated with the elliptic curve $y^2 = x^3 + A$, where $n$ is a natural number. The division polynomials are defined recursively, as described in the Wikipedia page on division polynomials.

I am working with the following expression:

$$\Psi_{2a} \Psi_b^4 - \Psi_{2b} \Psi_a^4$$ where $a$ and $b$ are natural numbers, and $a$ is odd. I know from the recursive definition of the division polynomials that the product $\Psi_a \Psi_b$ is a factor of this expression. However, using Sage Math for small values of $a$ and $b$, I observe that $\Psi_{b-a}$ also appears to be a factor (assuming $b > a$, without loss of generality).

I am trying to prove this factorization $\Psi_{2a} \Psi_b^4 - \Psi_{2b} \Psi_a^4$ contains $\Psi_{b-a}$ as factors, but I have been unable to do so through an argument, even after attempting an inductive approach on $(b-a)$.

Could anyone provide hints on how to approach this problem? In particular, I would appreciate suggestions on:

  1. The possible use of recursion relations or symmetry in the division polynomials to explain the factorization.
  2. How to structure an inductive proof for the factorization, or if there are alternative approaches.
  3. Any relevant references discussing the factorization of similar symmetric polynomials, especially in the context of division polynomials for elliptic curves.

EDIT: I am trying to show in general that whenever $a \equiv 1 \pmod 6$ and $b \equiv 5 \pmod 6$ then the expression has $\psi_{b-a}$ has a factor using induction on $b-a$. But the thing is getting more complicated whenever I am going to use the identity mentioned in the first answer.

Let $a=6k+1$ and $b=6k'+5$ so $b-a=6(k'-k)+4$. I need to show $\psi_{2a}\psi_b^4-\psi_{2a}\psi_a^4$ has factor $\psi_{b-a}.$

Here is my try:

$$\psi_{2a}\psi_b^4-\psi_{2a}\psi_a^4$$ $$=\psi_{2(6k+1)}\psi_{6k'+5}^4-\psi_{2(6k'+5)}\psi_{6k+1}^4$$ $$=\frac{\psi_{6k+1}}{2y}\left[\psi_{6k+3}\psi_{6k}^2-\psi_{6k-1}\psi_{6k+2}^2\right]\psi_{6k'+5}^4-\frac{\psi_{6k'+5}}{2y}\left[\psi_{6k'+7}\psi_{6k'+4}^2-\psi_{6k'+3}\psi_{6k'+6}^2\right]\psi_{6k+1}^4$$ $$= \frac{\psi_{6k+1}\psi_{6k'+5}}{2y}\left[\psi_{6k+3}\psi_{6k}^2\psi_{6k'+5}^3-\psi_{6k-1}\psi_{6k+2}^2\psi_{6k'+5}^3-\psi_{6k'+7}\psi_{6k'+4}^2\psi_{6k+1}^3+ \psi_{6k'+3}\psi_{6k'+6}^2\psi_{6k+1}^3 \right]$$

Now observe the terms inside the bracket. The difference of the indices in 1st and 3rd term is $b-a=6(k'-k)+4$. Similarly for 2nd and 4th term. But not able to proceed how to get $\psi_{6(k'-k)+4}$ outside.

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  • $\begingroup$ I took $x=1$, $y=2$, $y^2=x^3+3$, $a=1$, $b=4$ and got the ratio $(\Psi_{2a} \Psi_b^3 - \Psi_{2b} \Psi_a^4)/\Psi_{b-a}=1199/4. $ $\endgroup$ Commented Nov 11, 2024 at 17:04
  • $\begingroup$ I took $ x = -1, y = 1, y^2 = x^3 + 2, a = 1, b =5 $ and got the ratio $(\Psi_{2a} \Psi_b^3 - \Psi_{2b} \Psi_a^4)/\Psi_{b-a}=4284708659092920/71.$. $\endgroup$ Commented Nov 17, 2024 at 19:24
  • $\begingroup$ It should be $\psi_b^4$ instead of $\psi_b^3$. $\endgroup$ Commented Nov 18, 2024 at 4:02

1 Answer 1

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The most general identity for division polynomials is $$ \psi_{a+b}\psi_{a-b}\psi_{c+d}\psi_{c-d}=\psi_{a+c}\psi_{a-c}\psi_{b+d}\psi_{b-d}-\psi_{b+c}\psi_{b-c}\psi_{a+d}\psi_{a-d},$$ where all $a,b,c,d$ are integers or half-integers. It is equivalent to three-term identity for Weierstrass sigma-function $$\sigma(a+b)\sigma(a-b)\sigma(c+d)\sigma(c-d)-\sigma(a+c)\sigma(a-c)\sigma(b+d)\sigma(b-d)+\\ +\sigma(a+d)\sigma(a-d)\sigma(b+c)\sigma(b-c)=0.$$ All other identities should follow from this one because don't have other formulae including Weierstrass sigma-function only.

It looks like that this identity doesn't work here. Probably the conjecture is wrong. At least according to my calculations for $a=1$ and $b=4$ $$\Psi_{2a} \Psi_b^4 - \Psi_{2b} \Psi_a^4=72 x^3 y \left(27 x^6-36 x^3 y^2+8 y^4\right) \left(137674647 x^{21}-904671504 x^{18} y^2+2470881024 x^{15} y^4-3588803712 x^{12} y^6+2920093440 x^9 y^8-1257320448 x^6 y^{10}+219475968 x^3 y^{12}+2670592 y^{14}\right)$$ while $$\Psi_{b-a}=-3 x \left(11 x^3-12 y^2\right).$$

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  • $\begingroup$ Thanks for the identity. The expression is $\psi_{2a}\psi_{b}^4-\psi_{2b}\psi_{a}^4$ $\endgroup$ Commented Nov 11, 2024 at 17:43
  • $\begingroup$ One more thing the above identity is true for $a>b>c>d$. Or any other condition on $a,b,c,d$? $\endgroup$ Commented Nov 11, 2024 at 17:56
  • $\begingroup$ @DEBAJYOTIDE It was just misprint, formula is correct. No conditions on $a,b,c,d$, arbitrary integers. $\endgroup$ Commented Nov 11, 2024 at 17:58
  • $\begingroup$ However, the division polynomial is only defined for non-negative integers. If we choose $a<b$ it raises the question of how the left-hand side would be defined. It seems we might need to permute $a,b,c,d$ to ensure that all resulting indices are non-negative. $\endgroup$ Commented Nov 11, 2024 at 18:03
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    $\begingroup$ @DEBAJYOTIDE If you take the formula $\Psi_1^2\Psi_{n+k}\Psi_{n-k}=\Psi_{k}^2\Psi_{n+1}\Psi_{n-1}-\Psi_{k+1}\Psi_{k-1}\Psi_{n}^2$ and put $k=0$, $n=1$ you'll get $\Psi_{-1}=-\Psi_{1}.$ After that $n=0$ gives you $\Psi_{-n}=-\Psi_{n}.$ $\endgroup$ Commented Nov 11, 2024 at 18:16

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